Analysis of Properties of Function Involving Cosine of Reciprocal

To determine the correct options, let us analyze the properties of the function $f(x) = x \cos \left(\frac{1}{x}\right)$ for $x \ge 1$.

1. First Derivative of $f(x)$

Differentiating $f(x)$ with respect to $x$ using the product rule:
$f'(x) = \frac{d}{dx} \left[ x \cos \left(\frac{1}{x}\right) \right] = \cos\left(\frac{1}{x}\right) + x \left(-\sin\left(\frac{1}{x}\right)\right) \left(-\frac{1}{x^2}\right)$
$f'(x) = \cos\left(\frac{1}{x}\right) + \frac{1}{x}\sin\left(\frac{1}{x}\right)$

2. Evaluation of Option (B)

Taking the limit as $x \to \infty$:
$\lim_{x \to \infty} f'(x) = \lim_{x \to \infty} \left[ \cos\left(\frac{1}{x}\right) + \frac{1}{x}\sin\left(\frac{1}{x}\right) \right]$
As $x \to \infty$, $\frac{1}{x} \to 0$, which gives:
$\lim_{x \to \infty} f'(x) = \cos(0) + 0 \cdot \sin(0) = 1 + 0 = 1$
Thus, Option (B) is correct.

3. Monotonicity of $f'(x)$ (Option D)

To determine the monotonicity of $f'(x)$, we differentiate $f'(x)$ with respect to $x$:
$f''(x) = \frac{d}{dx} \left[ \cos\left(\frac{1}{x}\right) + \frac{1}{x}\sin\left(\frac{1}{x}\right) \right]$
$f''(x) = -\sin\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right) - \frac{1}{x^2}\sin\left(\frac{1}{x}\right) + \frac{1}{x}\cos\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)$
$f''(x) = \frac{1}{x^2}\sin\left(\frac{1}{x}\right) - \frac{1}{x^2}\sin\left(\frac{1}{x}\right) - \frac{1}{x^3}\cos\left(\frac{1}{x}\right) = -\frac{1}{x^3}\cos\left(\frac{1}{x}\right)$

For $x \in [1, \infty)$, we have $\frac{1}{x} \in (0, 1]$. In this interval, $\cos\left(\frac{1}{x}\right) > 0$ and $x^3 > 0$, hence:
$f''(x) = -\frac{1}{x^3}\cos\left(\frac{1}{x}\right) < 0$
Since $f''(x) < 0$ for all $x \in [1, \infty)$, $f'(x)$ is strictly decreasing in the interval $[1, \infty)$.
Thus, Option (D) is correct.

4. Evaluation of Options (A) and (C)

By Lagrange's Mean Value Theorem, since $f(x)$ is continuous on $[x, x+2]$ and differentiable on $(x, x+2)$ for any $x \ge 1$, there exists some $c \in (x, x+2)$ such that:
$\frac{f(x+2) - f(x)}{(x+2) - x} = f'(c) \implies f(x+2) - f(x) = 2f'(c)$

Since $f'(x)$ is strictly decreasing on $[1, \infty)$ and $\lim_{x \to \infty} f'(x) = 1$, the infimum of $f'(x)$ on $[1, \infty)$ is $1$. Therefore:
$f'(c) > 1 \quad \text{for all } c \in [1, \infty)$

Consequently:
$f(x+2) - f(x) = 2f'(c) > 2(1) = 2 \quad \text{for all } x \in [1, \infty)$

Hence, $f(x+2) - f(x) > 2$ holds for all $x \in [1, \infty)$.
Thus, Option (C) is correct, and Option (A) is incorrect.

Conclusion

The correct options are B, C, and D.